At the max the Queen can move in about 28 squares for choices which can be selected with the joysticks ..
one joystick for the pawns and one for the other major pieces ...!!!!!
the king has 8 choices
and the knight has 8 choices...
whereas at the most the bishop has 14 each
the rooks have 14 each too
And the pawns have 4 choices at the most taking into considerations the pawn jump in the start and enpassant
So now we have when pieces are centralized >
Using this theory we can always know how many choices we have to make depending on the pieces availaible the upper limit and the lower limit depends on the position when we add overshadowing ,pinning etc to it ..
2 bishops or 2 rooks equal one Queen or vice versa
and 2 knights equal 8 pawns or vice versa
King can be considered as a separate piece
OR
one king equals 2 pawns or a single knight
we can go on analyzing further
as such
where
1 rook or 1 bishop + 1 knight + 1.5 pawns= one queen
2 knights + 2.5 pawns = one queen
and so on and so forth
1 queen + 1 pawn = 8 pawns >> 1 queen = 7 pawns
1 rook or 1 bishop + 3.5 pawns = 1 queen = 7 pawns
1 king + 2 knight + 1 pawn = 1 queen = 7 pawns
1 king + 1 rook or 1 bishop `1.5 = 1 queen = 7 pawns
1 king + 1 knight + 3 pawns = 1 queen = 7 pawns
from a strictly theoretical reductionist point of view.
I am afraid your theory is not confirmed by practice.
If you set up an analysis board with 6 pieces or less you will see the exact outcome.
Can you elucidate please ?
So when taking the reductionist point of view for the absolute strength of the pieces we have the king which gets left out of the
weighing scale while when we take the upper limit of a complete board the queen gets left out as
2 knights + 8 pawns + 1 king = 2 rooks + 2 bishops = 2 queens
where we can substitute as follows
16c + 32c + 8c = 28c + 28c = 2 * 28c = 56c(choices when centralized)
so upper limit = 56c + 56c + 28c = 140c's for each opponent.
Getting the exact same outcome with fewer pieces confirms it ...;)
First the King needs no value, as it is always present.
A Queen is worth about two rooks, that is correct, well it is slightly less.
A rook is worth more than a bishop, about 2 pawns. An ending of rook vs. bishop + pawn is generally won for the rook, and an ending of rook vs. bishop + 2 pawns is usually drawn or won for the pawns.
A bishop is slightly worth more than a knight. Most endings of Bishop + pawn vs. knight + pawn are either won for the bishop or drawn.
A knight is generally worth somewhat less than 3 pawns. Endings of knight vs. 2 pawns are generally drawn. Endgames of knight versus 3 pawns are generally won for the pawns.
Now what we need to understand is overshadowing where the pieces are never in their full strength unless they are in the four central key squares ...
Thus we have
6 different pieces
1 knight
1 bishop
1 pawn
1 king
1 queen
1 rook
so taking
Ncr we have 15 combinations of the 6 pieces that can be arranged in the central key squares which helps us in reducing the 140c's to fewer choices and studying the relative position of the pieces helps us in calculating the number of choices that we can make the upper limit at every given position ...
But anyway it is not infinite atleast so it is good to know :)
Hey TPR im not talking about the traditional system ...where we give 9 points to a queen 5 to a rook
3 to a bishop and knight etc ...
im talking in terms of the choices or the maximum number of squares that we can choose to place a piece when in its full strength.
Practice has shown
P = 1
N = 2.8
B = 3
R = 5
Q = 9.5
You arrive at
P = 1
N = 2.5
B = 3.5
R = 3.5
Q = 7
This is wrong as explained by evidence from endgames above.